Complete Solutions Manual to accompany Chemical Principles Sixth Edition Steven S. Zumdahl Thomas J. Hummel Steven S. Zumdahl University of Illinois at Urbana-Champaign HOUGHTON MIFFLIN COMPANY Boston New York Vice President and Publisher: Charles Hartford Senior Marketing Manager: Nicole Moore Senior Development Editor: Rebecca Berardy Schwartz Supplements Editor: Kathryn White Editorial Associate: Henry Cheek Ancillary Coordinator: Sean McGann Marketing Associate: Kris Bishop Copyright 2009 by Houghton Mifflin Company. All rights reserved. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system without the prior written permission of Houghton Mifflin Company unless such copying is expressly permitted by federal copyright law. Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston, MA 02116. 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ISBN-13: 978-0-618-94822-2 ISBN-10: 0-618-94822-8 TABLE OF CONTENTS Page How to Use This Guide....................................................................................................................v Chapter 2 Atoms, Molecules, and Ions.....................................................................................1 Chapter 3 Stoichiometry.........................................................................................................16 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry ...................................51 Chapter 5 Gases......................................................................................................................96 Chapter 6 Chemical Equilibrium..........................................................................................149 Chapter 7 Acids and Bases ...................................................................................................192 Chapter 8 Applications of Aqueous Equilibria.....................................................................254 Chapter 9 Energy, Enthalpy, and Thermochemisty..............................................................341 Chapter 10 Spontaneity, Entropy, and Free Energy...............................................................369 Chapter 11 Electrochemistry ..................................................................................................413 Chapter 12 Quantum Mechanics and Atomic Theory............................................................455 Chapter 13 Bonding: General Concepts .................................................................................490 Chapter 14 Covalent Bonding: Orbitals ................................................................................540 Chapter 15 Chemical Kinetics................................................................................................576 Chapter 16 Liquids and Solids................................................................................................624 Chapter 17 Properties of Solutions.........................................................................................668 Chapter 18 The Representative Elements...............................................................................707 Chapter 19 Transition Metals and Coordination Chemistry...................................................730 Chapter 20 The Nucleus: A Chemists View.........................................................................756 Chapter 21 Organic Chemistry...............................................................................................773 iii v HOW TO USE THIS GUIDE Solutions to all of the end of chapter exercises are in this manual. This Solutions Guide can be very valuable if you use it properly. The way NOT to use it is to look at an exercise in the book and then immediately check the solution, often saying to yourself, Thats easy, I can do it. Developing problem solving skills takes practice. Dont look up a solution to a problem until you have tried to work it on your own. If you are completely stuck, see if you can find a similar problem in the Sample Exercises in the chapter. Only look up the solution as a last resort. If you do this for a problem, look for a similar problem in the end of chapter exercises and try working it. The more problems you do, the easier chemistry becomes. It is also in your self interest to try to work as many problems as possible. Most exams that you will take in chemistry will involve a lot of problem solving. If you have worked several problems similar to the ones on an exam, you will do much better than if the exam is the first time you try to solve a particular type of problem. No matter how much you read and study the text, or how well you think you understand the material, you dont really understand it until you have taken the information in the text and applied the principles to problem solving. You will make mistakes, but the good students learn from their mistakes. In this manual we have worked problems as in the textbook. We have shown intermediate answers to the correct number of significant figures and used the rounded answer in later calculations. Thus, some of your answers may differ slightly from ours. When we have not followed this convention, we have usually noted this in the solution. The most common exception is when working with the natural logarithm (ln) function, where we usually carried extra significant figures in order to reduce round-off error. In addition, we tried to use constants and conversion factors reported to at least one more significant figure as compared to numbers given in the problem. For some problems, this required the use of more precise atomic masses for H, C, N, and O as given in Chapter 3. This practice of carrying one extra significant figure in constants helps minimize round-off error. We are grateful to Claire Szoke for her outstanding effort in preparing the manuscript of this manual. We also thank Estelle Lebeau for her careful accuracy review and Jim Madru for his thorough copyediting of the Solutions Manual. We also are grateful to Don DeCoste for his assistance in creating solutions to some of the problems in this Solutions Guide. TJH SSZ 1 CHAPTER 2 ATOMS, MOLECULES, AND IONS Development of the Atomic Theory 18. Law of conservation of mass: mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g hydrogen for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers. For CO2 and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio. 19. From Avogadros hypothesis (law), volume ratios are equal to molecule ratios at constant temperature and pressure. Therefore, we can write a balanced equation using the volume data, Cl2 + 3 F2 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF3 for a balanced reaction. 20. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadros hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H2 + Cl2 2 HCl. From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted. 21. Hydrazine: 1.44 110 g H/g N; Ammonia: 2.16 110g H/g N; Hydrogen azide: 2.40 210g H/g N; Let's try all of the ratios: 0240 . 0144 . 0 = 6.00; 0240 . 0216 . 0 = 9.00; 144 . 0216 . 0 = 1.50 = 23 All the masses of hydrogen in these three compounds can be expressed as simple whole- number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6 : 9 : 1. 2 CHAPTER 2 ATOMS, MOLECULES, AND IONS 22. The law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1: 27.2 g C and 72.8 g O (100.0 - 27.2 = mass O) Compound 2: 42.9 g C and 57.1 g O (100.0 - 42.9 = mass O) The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: O g 8 . 72C g 2 . 27 = 0.374 g C/g O Compound 2: O g 1 . 57C g 9 . 42 = 0.751 g C/g O 12374 . 0751 . 0= ; this supports the law of multiple proportions as this carbon ratio is a whole number. 23. To get the atomic mass of H to be 1.00, we divide the mass that reacts with 1.00 g of oxygen by 0.126, that is, 0.126/0.126 = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na: 126 . 0875 . 2 = 22.8; Mg: 126 . 0500 . 1 = 11.9; O: 126 . 000 . 1 = 7.94 H O Na Mg Relative value 1.00 7.94 22.8 11.9 Accepted value 1.0079 15.999 22.99 24.31 The atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close. Something must be wrong about the assumed formulas of the compounds. It turns out that the correct formulas are H2O, Na2O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H. The Nature of the Atom 24. Deflection of cathode rays by magnetic and electrical fields led to the conclusion that they were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electrical field. particles are electrons. A cathode ray is a stream of electrons ( particles). 25. Density of hydrogen nucleus (contains one proton only): CHAPTER 2 ATOMS, MOLECULES, AND IONS 3 Vnucleus = 3 40 3 14 3cm 10 5 ) cm 10 5 ( ) 14 . 3 (34r a34 = = d = density = 3 153 4024g/cm 10 3cm 10 5g 10 67 . 1 = Density of H atom (contains one proton and one electron): Vatom = 3 24 3 8cm 10 4 ) cm 10 1 ( ) 14 . 3 (34 = d = 33 2428 24g/cm 0.4cm 10 4g 10 9 g 10 67 . 1= + 26. From Fig. 2.13 of the text, the average diameter of the nucleus is , cm 1013 ~ and the average diameter of the volume where the electrons roam about is . cm 108 ~ cm 10cm 10138 = 105; grape 1in 360 , 63grape 1ft 5280grape 1mile 1= = Because the grape needs to be 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(1 105) ~ 0.6 in. This is a reasonable size for a grape. 27. First, divide all charges by the smallest quantity, 6.40 1013. 131210 40 . 610 56 . 2 = 4.00; 640 . 068 . 7 = 12.00; 640 . 084 . 3 = 6.00 Because all charges are whole-number multiples of 6.40 1013 zirkombs, the charge on one electron could be 6.40 1013 zirkombs. However, 6.40 1013 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 1013 zirkombs or an integer fraction of 6.40 1013. 28. The proton and neutron have similar mass, with the mass of the neutron slightly larger than that of the proton. Each of these particles has a mass approximately 1800 times greater than that of an electron. The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom. 29. If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then o particles should have traveled through the thin foil with very minor deflections in their path. This was not the case because a few of the o particles were deflected at very large angles. Rutherford reasoned that the large deflections of these o particles could be caused only by a center of concentrated positive charge that contains most of the atoms mass (the nuclear model of the atom). 4 CHAPTER 2 ATOMS, MOLECULES, AND IONS Elements and the Periodic Table 30. a. A molecule has no overall charge (an equal number of electrons and protons are present). Ions, on the other hand, have electrons added to form anions (negatively charged ions) or electrons removed to form cations (positively charged ions). b. The sharing of electrons between atoms is a covalent bond. An ionic bond is the force of attraction between two oppositely charged ions. c. A molecule is a collection of atoms held together by covalent bonds. A compound is composed of two or more different elements having constant composition. Covalent and/or ionic bonds can hold the atoms together in a compound. Another difference is that molecules do not necessarily have to be compounds. H2 is two hydrogen atoms held together by a covalent bond. H2 is a molecule, but it is not a compound; H2 is a diatomic element. d. An anion is a negatively charged ion, for example, Cl, O2, and SO42 are all anions. A cation is a positively charged ion, for example, Na+, Fe3+, and NH4+ are all cations. 31. The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter 3, the average mass of an atom is taken from a measurement made on a large number of atoms. The average atomic mass value is listed in the periodic table. 32. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am. Nonmetals: Si, B, At, Rn, and Br. b. Si, Ge, B, and At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. Aluminum has mostly properties of metals, so it is generally not classified as a metalloid. 33. a. The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon). Radon has only radioactive isotopes. In the periodic table, the whole number enclosed in parentheses is the mass number of the longest-lived isotope of the element. b. promethium (Pm) and technetium (Tc) 34. Carbon is a nonmetal. Silicon and germanium are called metalloids as they exhibit both metallic and nonmetallic properties. Tin and lead are metals. Thus metallic character increases as one goes down a family in the periodic table. The metallic character decreases from left to right across the periodic table. 35. a. Five: F, Cl, Br, I, and At b. Six: Li, Na, K, Rb, Cs, and Fr (H is not considered an alkali metal.) c. 14: Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, and Lu d. 40: All elements in the block defined by Sc, Zn, Uub, and Ac are transition metals. CHAPTER 2 ATOMS, MOLECULES, AND IONS 5 36. a. Cobalt is element 27. A = mass number = 27 + 31 = 58; 5827Co b. 105B c. 2312Mg d. 13253I e. 199F f. 6529Cu 37. a. 2412Mg: 12 protons, 12 neutrons, 12 electrons b. 2412Mg2+: 12 p, 12 n, 10 e c. 5927Co2+: 27 p, 32 n, 25 e d. 5927Co3+: 27 p, 32 n, 24 e e. 5927Co: 27 p, 32 n, 27 e f. 7934Se: 34 p, 45 n, 34 e g. 7934Se2: 34 p, 45 n, 36 e h. 6328Ni: 28 p, 35 n, 28 e i. 5928Ni2+: 28 p, 31 n, 26 e 38. Symbol Number of Protons in Nucleus Number of Neutrons inNucleus Number of Electrons Net Charge 23892U 92 146 92 0 4020Ca2+ 20 20 18 2+ 5123V3+ 23 28 20 3+ 8939Y 39 50 39 0 7935Br 35 44 36 1 3115P3 15 16 18 3 39. Atomic number = 63 (Eu); net charge = +63 60 = 3+; mass number = 63 + 88 = 151; symbol: 15163Eu3+ Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 48 = 2+; symbol: 11850Sn2+. 6 CHAPTER 2 ATOMS, MOLECULES, AND IONS 40. Atomic number = 16 (S); net charge = +16 18 = 2; mass number = 16 + 18 = 34; symbol: 3416S2 Atomic number = 16 (S); net charge = +16 18 = 2; Mass number = 16 + 16 = 32; symbol: 3216S2 41. In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A and 3A metals form stable 1+, 2+ and 3+ charged cations, respectively. Group 5A, 6A, and 7A nonmetals form 3, 2 and 1 charged anions, respectively. a. Lose 2 e to form Ra2+. b. Lose 3 e to form In3+. c. Gain 3 e to form 3P . d. Gain 2 e to form 2Te . e. Gain 1 e to form Br. f. Lose 1 e to form Rb+. 42. See Exercise 41 for a discussion of charges various elements form when in ionic compounds. a. Element 13 is Al. Al forms 3+ charged ions in ionic compounds. Al3+ b. Se2 c. Ba2+ d. N3 e. Fr+ f. Br Nomenclature 43. AlCl3, aluminum chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; AlCl3 and CrCl3 are ionic compounds following the rules for naming ionic compounds. The major difference is that CrCl3 contains a transition metal (Cr) that generally exhibits two or more stable charges when in ionic compounds. We need to indicate which charged ion we have in the compound. This is generally true whenever the metal in the ionic compound is a transition metal. ICl3 is made from only nonmetals and is a covalent compound. Predicting formulas for covalent compounds is extremely difficult. Because of this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the first species present in the formula; when this is the case, mono- is not used (it is assumed). 44. a. Dinitrogen monoxide is correct. N and O are both nonmetals resulting in a covalent compound. We need to use the covalent rules of nomenclature. The other two names are for ionic compounds. b. Copper(I) oxide is correct. With a metal in a compound, we have an ionic compound. Because copper, like most transition metals, forms at least a couple of different stable charged ions, we must indicate the charge on copper in the name. Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds. Dicopper monoxide is the name if this were a covalent compound, which it is not. c. Lithium oxide is correct. Lithium forms 1+ charged ions in stable ionic compounds. Because lithium is assumed to form 1+ ions in compounds, we do not need to indicate the charge of the metal ion in the compound. Dilithium monoxide would be the name if Li2O were a covalent compound (a compound composed of only nonmetals). CHAPTER 2 ATOMS, MOLECULES, AND IONS 7 45. a. sulfur difluoride b. dinitrogen tetroxide c. iodine trichloride d. tetraphosphorus hexoxide 46. a. sodium perchlorate b. magnesium phosphate c. aluminum sulfate d. sulfur difluoride e. sulfur hexafluoride f. sodium hydrogen phosphate g. sodium dihydrogen phosphate h. lithium nitride i. sodium hydroxide j. magnesium hydroxide k. aluminum hydroxide l. silver chromate 47. a. copper(I) iodide b. copper(II) iodide c. cobalt(II) iodide d. sodium carbonate e. sodium hydrogen carbonate or sodium bicarbonate f. tetrasulfur tetranitride g. sulfur hexafluoride h. sodium hypochlorite i. barium chromate j. ammonium nitrate 48. a. acetic acid b. ammonium nitrite c. colbalt(III) sulfide d. iodine monochloride e. lead(II) phosphate f. potassium iodate g. sulfuric acid h. strontium nitride i. aluminum sulfite j. tin(IV) oxide k. sodium chromate l. hypochlorous acid 49. a. SO2 b. SO3 c. Na2SO3 d. KHSO3 e. Li3N f. Cr2(CO3)3 g. Cr(C2H3O2)2 h. SnF4 i. NH4HSO4: composed of NH4+ and HSO4 ions j. (NH4)2HPO4 k. KClO4 l. NaH m. HBrO n. HBr 50. a. Na2O b. Na2O2 c. KCN d. Cu(NO3)2 e. SiCl4 f. PbO g. PbO2 h. CuCl i. GaAs: We would predict the stable ions to be Ga3+ and As3. j. CdSe k. ZnS l. Hg2Cl2: Mercury(I) exists as Hg22+. m. HNO2 n. P2O5 51. a. Pb(C2H3O2)2; lead(II) acetate b. CuSO4; copper(II) sulfate c. CaO; calcium oxide d. MgSO4; magnesium sulfate e. Mg(OH)2; magnesium hydroxide f. CaSO4; calcium sulfate g. N2O; dinitrogen monoxide or nitrous oxide (common name) 8 CHAPTER 2 ATOMS, MOLECULES, AND IONS 52. a. Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron. Iron(III) chloride is correct. b. This is a covalent compound so use the covalent rules. Nitrogen dioxide is correct. c. This is an ionic compound, so use the ionic rules. Calcium oxide is correct. Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed. d. This is an ionic compound, so use the ionic rules. Aluminum sulfide is correct. e. This is an ionic compound, so use the ionic rules. Mg is magnesium. Magnesium acetate is correct. f. Because phosphate has a 3 charge, the charge on iron is 3+. Iron(III) phosphate is correct. g. This is a covalent compound, so use the covalent rules. Diphosphorus pentasulfide is correct. h. Because each sodium is 1+ charged, we have the O22 (peroxide) ion present. Sodium peroxide is correct. Note that sodium oxide would be Na2O. i. HNO3 is nitric acid, not nitrate acid. Nitrate acid does not exist. j. H2S is hydrosulfuric acid or dihydrogen sulfide or just hydrogen sulfide (common name). H2SO4 is sulfuric acid. 53. a. nitric acid, HNO3 b. perchloric acid, HClO4 c. acetic acid, HC2H3O2 d. sulfuric acid, H2SO4 e. phosphoric acid, H3PO4 Additional Exercises 54. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons and neutrons, which can be broken down into quarks. For our purpose, electrons, neutrons, and protons are the key smaller parts of an atom. b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have 0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions were included, then different ions/atoms of H could have different numbers of electrons. c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2 protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then the number of electrons will be 1 for hydrogen and 2 for helium. CHAPTER 2 ATOMS, MOLECULES, AND IONS 9 d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1 g hydrogen for every 16 g of O present. These are distinctly different compounds, each with its own unique relative number and types of atoms present. e. A chemical equation involves a reorganization of the atoms. Bonds are broken between atoms in the reactants, and new bonds are formed in the products. The number and types of atoms between reactants and products do not change. Because atoms are conserved in a chemical reaction, mass is also conserved. 55. A compound will always have a constant composition by mass. From the initial data given, the mass ratio of H : S : O in sulfuric acid (H2SO4) is: 02 . 200 . 64:02 . 207 . 32:02 . 202 . 2 = 1 : 15.9 : 31.7 If we have 7.27 g H, then we will have 7.27 15.9 = 116 g S and 7.27 31.7 = 230. g O in the second sample of H2SO4. 56. Mass is conserved in a chemical reaction. Chromium(III) oxide + aluminum chromium + aluminum oxide Mass: 34.0 g 12.1 g 23.3 ? Mass aluminum oxide produced = (34.0 + 12.1) 23.3 = 22.8 g 57. From the Na2X formula, X has a 2 charge. Because 36 electrons are present, X has 34 protons, 79 34 = 45 neutrons, and is selenium. a. True. Nonmetals bond together using covalent bonds and are called covalent compounds. b. False. The isotope has 34 protons. c. False. The isotope has 45 neutrons. d. False. The identity is selenium, Se. 58. a. Fe2+: 26 protons (Fe is element 26.); protons electrons = charge, 26 2 = 24 electrons; FeO is the formula because the oxide ion has a 2 charge. b. Fe3+: 26 protons; 23 electrons; Fe2O3 c. Ba2+: 56 protons; 54 electrons; BaO d. Cs+: 55 protons; 54 electrons; Cs2O e. S2: 16 protons; 18 electrons; Al2S3 f. P3: 15 protons; 18 electrons; AlP g. Br: 35 protons; 36 electrons; AlBr3 h. N3: 7 protons; 10 electrons; AlN 59. From the XBr2 formula, the charge on element X is 2+. Therefore, the element has 88 protons, which identifies it as radium, Ra. 230 88 = 142 neutrons. 10 CHAPTER 2 ATOMS, MOLECULES, AND IONS 60. The solid residue must have come from the flask. 61. In the case of sulfur, SO42 is sulfate, and SO32 is sulfite. By analogy: SeO42: selenate; SeO32: selenite; TeO42: tellurate; TeO32: tellurite 62. The polyatomic ions and acids in this problem are not named in the text. However, they are all related to other ions and acids named in the text that contain a same group element. Because HClO4 is perchloric acid, HBrO4 would be perbromic acid. Because ClO3 is the chlorate ion, KIO3 would be potassium iodate. Since ClO2 is the chlorite ion, NaBrO2 would be sodium bromite. And finally, because HClO is hypochlorous acid, HIO would be hypo-iodous acid. 63. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or: O g 000 . 1In g 784 . 400 . 16A= , A = atomic mass of In = 76.54 If the formula is In2O3, then two times the atomic mass of In will combine with three times the atomic mass of O, or: O g 000 . 1In g 784 . 400 . 16 ) 3 (A 2= , A = atomic mass of In = 114.8 The latter number is the atomic mass of In used in the modern periodic table. 64. a. Ca2+ and N3: Ca3N2, calcium nitride b. K+ and O2: K2O, potassium oxide c. Rb+ and F: RbF, rubidium fluoride d. Mg2+ and S2: MgS, magnesium sulfide e. Ba2+ and I: BaI2, barium iodide f. Al3+ and Se2: Al2Se3, aluminum selenide g. Cs+ and P3: Cs3P, cesium phosphide h. In3+ and Br: InBr3, indium(III) bromide; In also forms In+ ions, but you would predict In3+ ions from its position in the periodic table. 65. The cation has 51 protons and 48 electrons. The number of protons corresponds to the atomic number. Thus this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1. The formula of the compound formed between Sb3+ and Cl is SbCl3. The name of the compound is antimony(III) chloride. The Roman numeral is used to indicate the charge of Sb because the predicted charge is not obvious from the periodic table. 66. a. This is element 52, tellurium. Te forms stable 2 charged ions in ionic compounds (like other oxygen family members). CHAPTER 2 ATOMS, MOLECULES, AND IONS 11 b. Rubidium. Rb, element 37, forms stable 1+ charged ions. c. Argon. Ar is element 18. d. Astatine. At is element 85. 67. Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present. The heavier elements have significantly more neutrons than protons in their nuclei. Because this element forms anions, it is a nonmetal and will be a halogen because halogens form stable 1 charged ions in ionic compounds. From the halogens listed, chlorine, with an average atomic mass of 35.45, fits the data. The two isotopes are 35Cl and 37Cl, and the number of electrons in the 1 ion is 18. Note that because the atomic mass of chlorine listed in the periodic table is closer to 35 than 37, we can assume that 35Cl is the more abundant isotope. This is discussed in Chapter 3. Challenge Problems 68. Because the gases are at the same temperature and pressure, the volumes are directly proportional to the number of molecules present. Lets consider hydrogen and oxygen to be monatomic gases and that water has the simplest possible formula (HO). We have the equation: H + O HO But the volume ratios are also equal to the molecule ratios, which correspond to the coefficients in the equation: 2 H + O 2 HO Because atoms cannot be created nor destroyed in a chemical reaction, this is not possible. To correct this, we can make oxygen a diatomic molecule: 2 H + O2 2 HO This does not require hydrogen to be diatomic. Of course, if we know water has the formula H2O, we get: 2 H + O2 2 H2O The only way to balance this is to make hydrogen diatomic: 2 H2 + O2 2 H2O 69. a. Both compounds have C2H6O as the formula. Because they have the same formula, their mass percent composition will be identical. However, these are different compounds with different properties because the atoms are bonded together differently. These compounds are called isomers of each other. b. When wood burns, most of the solid material in wood is converted to gases, which escape. The gases produced are most likely CO2 and H2O. c. The atom is not an indivisible particle but is instead composed of other smaller particles, for example, electrons, neutrons, and protons. 12 CHAPTER 2 ATOMS, MOLECULES, AND IONS d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although the compounds are composed of different isotopes, their properties are similar because different isotopes of the same element have similar properties (except, of course, their mass). 70. For each experiment, divide the larger number by the smaller. In doing so, we get: experiment 1 X = 1.0 Y = 10.5 experiment 2 Y = 1.4 Z = 1.0 experiment 3 X = 1.0 Y = 3.5 Our assumption about formulas dictates the rest of the solution. For example, if we assume that the formula of the compound in experiment 1 is XY and that of experiment 2 is YZ, we get relative masses of: X = 2.0; Y = 21; Z = 15 (= 21/1.4) and a formula of X3Y for experiment 3 [three times as much X must be present in experiment 3 as compared to experiment 1 (10.5/3.5 = 3)]. However, if we assume the formula for experiment 2 is YZ and that of experiment 3 is XZ, then we get: X = 2.0; Y = 7.0; Z = 5.0 (= 7.0/1.4) and a formula of XY3 for experiment 1. Any answer that is consistent with your initial assumptions is correct. The answer to part d depends on which (if any) of experiments 1 and 3 have a formula of XY in the compound. If the compound in expt. 1 has formula XY, then: 21 g XY XY g ) 4 . 0 2 . 4 (Y g 2 . 4+ = 19.2 g Y (and 1.8 g X) If the compound in experiment 3 has the XY formula, then: 21 g XY XY g ) 0 . 2 0 . 7 (Y g 0 . 7+ = 16.3 g Y (and 4.7 g X) Note that it could be that neither experiment 1 nor experiment 3 has XY as the formula. Therefore, there is no way of knowing an absolute answer here. 71. Compound I: Q g 00 . 3R g 0 . 14 = Q g 00 . 1R g 67 . 4; Compound II: Q g 50 . 4R g 00 . 7 = Q g 00 . 1R g 56 . 1 The ratio of the masses of R that combines with 1.00 g Q is 56 . 167 . 4 = 2.99 - 3. CHAPTER 2 ATOMS, MOLECULES, AND IONS 13 As expected from the law of multiple proportions, this ratio is a small whole number. Because compound I contains three times the mass of R per gram of Q as compared with compound II (RQ), the formula of compound I should be R3Q. 72. Most of the mass of the atom is due to the protons and the neutrons in the nucleus, and protons and neutrons have about the same mass (1.67 1024 g). The ratio of the mass of the molecule to the mass of a nuclear particle will give a good approximation of the number of nuclear particles (protons and neutrons) present. g 10 67 . 1g 10 31 . 72423 = 43.8 ~ 44 nuclear particles Thus there are 44 protons and neutrons present. If the number of protons equals the number of neutrons, we have 22 protons in the molecule. One possibility would be the molecule CO2 [6 + 2(8) = 22 protons]. 73. Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal numbers of molecules. In terms of balanced equations, Avogadros hypothesis (law) implies that volume ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule of CxHy produces 8 molecules of CO2 and 9 molecules of H2O. CxHy + n O2 8 CO2 + 9 H2O. Because all the carbon in octane ends up as carbon in CO2, octane must contain 8 atoms of C. Similarly, all hydrogen in octane ends up as hydrogen in H2O, so one molecule of octane must contain 9 2 = 18 atoms of H. Octane formula = C8H18 and the ratio of C:H = 8:18 or 4:9. 74. Let Xa be the formula for the atom/molecule X, Yb be the formula for the atom/molecule Y, XcYd be the formula of compound I between X and Y, and XeYf be the formula of compound II between X and Y. Using the volume data, the following would be the balanced equations for the production of the two compounds. Xa + 2 Yb 2 XcYd; 2 Xa + Yb 2 XeYf From the balanced equations, a = 2c = e and b = d = 2f. Substituting into the balanced equations: X2c + 2 Y2f 2 XcY2f 2 X2c + Y2f 2 X2cYf For simplest formulas, assume that c = f = 1. Thus: X2 + 2 Y2 2 XY2 and 2 X2 + Y2 2 X2Y Compound I = XY2: If X has relative mass of 1.00, y 2 00 . 100 . 1+= 0.3043, y = 1.14. 14 CHAPTER 2 ATOMS, MOLECULES, AND IONS Compound II = X2Y: If X has relative mass of 1.00, y + 00 . 200 . 2= 0.6364, y = 1.14. The relative mass of Y is 1.14 times that of X. Thus if X has an atomic mass of 100, then Y will have an atomic mass of 114. Marathon Problem 75. a. For each set of data, divide the larger number by the smaller number to determine relative masses. 295 . 0602 . 0 = 2.04; A = 2.04 when B = 1.00 172 . 0401 . 0 = 2.33; C = 2.33 when B = 1.00 320 . 0374 . 0 = 1.17; C = 1.17 when A = 1.00 To have whole numbers, multiply the results by 3. Data set 1: A = 6.1 and B = 3.0 Data set 2: C = 7.0 and B = 3.0 Data set 3: C = 3.5 and A = 3.0 or C = 7.0 and A = 6.0 Assuming 6.0 for the relative mass of A, the relative masses would be A = 6.0, B = 3.0, and C = 7.0 (if simplest formulas are assumed). b. Gas volumes are proportional to the number of molecules present. There are many possible correct answers for the balanced equations. One such solution that fits the gas volume data is: 6 A2 + B4 4 A3B B4 + 4 C3 4 BC3 3 A2 + 2 C3 6 AC In any correct set of reactions, the calculated mass data must match the mass data given initially in the problem. Here, the new table of relative masses would be: 295 . 0602 . 0B mass) A mass ( 642= ; mass A2 = 0.340(mass B4) 172 . 0401 . 0B mass) C mass ( 443= ; mass C3 = 0.583(mass B4) 320 . 0374 . 0) A mass ( 3) C mass ( 223= ; mass A2 = 0.570(mass C3) CHAPTER 2 ATOMS, MOLECULES, AND IONS 15 Assume some relative mass number for any of the masses. We will assume that mass B = 3.0, so mass B4 = 4(3.0) = 12. Mass C3 = 0.583(12) = 7.0, mass C = 7.0/3 Mass A2 = 0.570(7.0) = 4.0, mass A = 4.0/2 = 2.0 When we assume a relative mass for B = 3.0, then A = 2.0 and C = 7.0/3. The relative masses having all whole numbers would be A = 6.0, B = 9.0, and C = 7.0. Note that any set of balanced reactions that confirms the initial mass data is correct. This is just one possibility. 16 CHAPTER 3 STOICHIOMETRY Atomic Masses and the Mass Spectrometer 23. A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium). 24. Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 x. 151.96 = 100) 9209 . 152 )( 100 ( ) 9196 . 150 ( x x + 15196 = (150.9196)x + 15292.09 (152.9209)x, 96 = (2.0013)x x = 48%; 48% 151Eu and 100 48 = 52% 153Eu 25. 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 - 117.0 = 0.3740(A) A = 3740 . 02 . 69 = 185 amu (A = 184.95 amu without rounding to proper significant figures) 26. A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766) A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 amu; from the periodic table, the element is Pb. 27. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes, differing in mass by two mass units. The peak at 157.84 corresponds to a Br2 molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2, or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to 161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present at about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br. CHAPTER 3 STOICHIOMETRY 17 28. Scaled Intensity Compound Mass Intensity Largest Peak = 100 __________________________________________________________________ H2120Te 121.92 0.09 0.3 H2122Te 123.92 2.46 7.1 H2123Te 124.92 0.87 2.5 H2124Te 125.92 4.61 13.4 H2125Te 126.92 6.99 20.3 H2126Te 127.92 18.71 54.3 H2128Te 129.92 31.79 92.2 H2130Te 131.93 34.48 100.0 29. GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have two peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2. Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have three peaks at 288, 290, and 292 with intensities in the ratio of 36 : 48 : 16 or 9 : 12 : 4. We get this ratio from the following probability table: 100500122 124 126 128 130 132 134 144 14618 CHAPTER 3 STOICHIOMETRY 69Ga (0.60) 71Ga (0.40) 69Ga (0.60) 0.36 0.24 71Ga (0.40) 0.24 0.16 288 290 292 Moles and Molar Masses 30. 4.24 g C6H6 g 11 . 78mol 1 = 5.43 210 mol C6H6 5.43 210 mol C6H6 molmolecules 10 022 . 623 = 3.27 1022 molecules C6H6 Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 total atoms. 3.27 1022 molecules C6H6 moleculetotal atoms 12 = 3.92 1023 atoms total 0.224 mol H2O molg 02 . 18 = 4.04 g H2O 0.224 mol H2O molmolecules 10 022 . 623 = 1.35 1023 molecules H2O 1.35 1023 molecules H2O moleculetotal atoms 3 = 4.05 1023 atoms total 2.71 1022 molecules CO2 molecules 10 022 . 6mol 123 = 4.50 210 mol CO2 4.50 210 mol CO2 molg 01 . 44 = 1.98 g CO2 2.71 1022 molecules CO2 2CO moleculetotal atoms 3 = 8.13 1022 atoms total 3.35 1022 atoms total total atoms 6molecule 1 = 5.58 1021 molecules CH3OH CHAPTER 3 STOICHIOMETRY 19 5.58 1021 molecules CH3OH molecules 10 022 . 6mol 123 = 9.27 310 mol CH3OH 9.27 310 mol CH3OH molg 04 . 32 = 0.297 g CH3OH 31. a. 20.0 mg C8H10N4O2 g 20 . 194mol 1mg 1000g 1 = 1.03 410mol C8H10N4O2 b. 2.72 1021 molecules C2H5OH molecules 10 022 . 6mol 123 = 4.52 310mol C2H5OH c. 1.50 g CO2 g 01 . 44mol 1= 3.41 210mol CO2 32. a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show how these conversion factors can be used. Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol 5.00 g C2H5O2N N O H C molN O H C molecules 10 022 . 6N O H C g 07 . 75N O H C mol 12 5 22 5 2232 5 22 5 2 N O H C moleculeN atom 12 5 2 = 4.01 1022 atoms N b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 5.00 g Mg3N2 2 32 3232 32 3N Mg molN Mg units formula 10 022 . 6N Mg g 95 . 100N Mg mol 1 2 3N Mg molatoms 2 = 5.97 1022 atoms N c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol 5.00 g Ca(NO3)2 2 32 3) NO ( Ca g 10 . 164) NO ( Ca mol 12 3) NO ( Ca molN mol 2 N molN atoms 10 022 . 623 = 3.67 1022 atoms N 20 CHAPTER 3 STOICHIOMETRY d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4 4 2 4 24 2O N molN mol 2O N g 02 . 92O N mol 1N molN atoms 10 022 . 623 = 6.54 1022 atoms N 33. 4.0 g H2 H mol 1H atoms 10 022 . 6H mol 1H mol 2H g 016 . 2H mol 1232 22 = 2.4 1024 atoms 4.0 g He He mol 1He atoms 10 022 . 6He g 003 . 4He mol 123 = 6.0 1023 atoms 1.0 mol F2 F mol 1F atoms 10 022 . 6F mol 1F mol 2232 = 1.2 1024 atoms 44.0 g CO2 atoms mol 1atoms 10 022 . 6CO mol 1) O 2 C 1 ( atoms mol 3CO g 01 . 44CO mol 1232 22 + = 1.81 1024 atoms 146 g SF6 atoms mol 1atoms 10 022 . 6SF mol 1) F 6 S 1 ( atoms mol 7SF g 07 . 146SF mol 1236 66 + = 4.21 1024 atoms The order is: 4.0 g He < 1.0 mol F2 < 44.0 g CO2 < 4.0 g H2 < 146 g SF6 34. a. 14 mol C C molg 011 . 12+ 18 mol H H molg 0079 . 1 + 2 mol N N molg 007 . 14 + 5 mol O O molg 999 . 15 = 294.305 g/mol b. 10.0 g C14H18N2O5 5 2 18 145 2 18 14O N H C g 3 . 294O N H C mol 1 = 3.40 102 mol C14H18N2O5 c. 1.56 mol molg 3 . 294 = 459 g C14H18N2O5 d. 5.0 mg molmolecles 10 02 . 6g 3 . 294mol 1mg 1000g 123 = 1.0 1019 molecules C14H18N2O5 e. 1.2 g C14H18N2O5 N molN atoms 10 02 . 6O N H C molN mol 2O N H C g 3 . 294O N H C mol 1235 2 18 14 5 2 18 145 2 18 14 = 4.9 1021 atoms N CHAPTER 3 STOICHIOMETRY 21 f. 1.0 109 molecules molg 3 . 294atoms 10 02 . 6mol 123 = 4.9 1013 g g. 1 molecule molg 305 . 294atoms 10 022 . 6mol 123 = 4.887 1022 g C14H18N2O5 35. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol b. 500.0 g g 39 . 165mol 1 = 3.023 mol C2H3Cl3O2 c. 2.0 10-2 mol molg 39 . 165 = 3.3 g C2H3Cl3O2 d. 5.0 g C2H3Cl3O2 moleculeCl atoms 3molmolecules 10 02 . 6g 39 . 165mol 123 = 5.5 1022 atoms of chlorine e. 1.0 g Cl 2 3 3 22 3 3 2 2 3 3 2O Cl H C molO Cl H C g 39 . 165Cl mol 3O Cl H C mol 1g 45 . 35Cl mol 1 = 1.6 g chloral hydrate f. 500 molecules molg 39 . 165molecules 10 022 . 6mol 123 = 1.373 1019 g 36. 1.0 lb flour 2 4 22 4 2 2 4 29Br H C g 9 . 187Br H C mol 1flour gBr H C g 10 0 . 30flour lbflour g 454 2 4 223Br H C molmolecules 10 02 . 6 = 4.4 1016 molecules C2H4Br2 Percent Composition 37. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol Mass % C = compound g 43 . 336C g ) 01 . 12 ( 20 100 = 71.40% C Mass % H = compound g 43 . 336H g ) 008 . 1 ( 29 100 = 8.689% H 22 CHAPTER 3 STOICHIOMETRY Mass % F = compound g 43 . 336F g 00 . 19 100 = 5.648% F Mass % O = 100.00 (71.40 + 8.689 + 5.648) = 14.26% O or: Mass % O = compound g 43 . 336O g ) 00 . 16 ( 3 100 = 14.27% O 38. a. C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C = 2 4 10 8O N H C g 20 . 194C g ) 01 . 12 ( 8 100 = 20 . 19408 . 96 100 = 49.47% C b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = 11 22 2 1O H C g 30 . 342C g ) 01 . 12 ( 12 100 = 42.10% C c. C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = OH H C g 07 . 46C g ) 01 . 12 ( 25 2 100 = 52.14% C The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 39. For each compound, determine the mass of 1 mole of compound, and then calculate the mass percentage of N in 1 mole of that compound. a. NO: % N = NO g 006 . 30N g 007 . 14 100 = 46.681% N b. NO2: % N = 2NO g 005 . 46N g 007 . 14 100 = 30.447% N c. N2O4: % N = 4 2O N g 010 . 92N g 014 . 28 100 = 30.447% N d. N2O: % N = O N g 013 . 44N g 014 . 282 100 = 63.649% N 40. Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co Co molamin cyanocobal mol 1Co g 58.93Co mol 1 = 7.36 102 mol cyanocobalamin CHAPTER 3 STOICHIOMETRY 23 mol 10 36 . 7g 00 . 100amin cyanocobal mol 1amin cyanocobal g2 =x, x = molar mass = 1360 g/mol 41. There are 0.390 g Cu for every 100.000 g of fungal laccase. Lets assume 100.000 g fungal laccase. Mol fungal laccase = 0.390 g Cu Cu mol 4laccase fungal mol 1Cu g 63.55Cu mol 1 = 1.53 103 mol laccase fungal mol 1laccase fungal g x = mol 10 53 . 1g 000 . 1003 , x = molar mass = 6.54 104 g/mol 42. If we have 100.0 g of Portland cement, we have 50. g Ca3SiO5, 25 g Ca2SiO4, 12 g Ca3Al2O6, 8.0 g Ca2AlFeO5, and 3.5 g CaSO4-2H2O. Mass percent Ca: 50. g Ca3SiO5 Ca mol 1Ca g 08 . 40SiO Ca mol 1Ca mol 3SiO Ca g 33 . 228SiO Ca mol 15 3 5 35 3 = 26 g Ca 25 g Ca2SiO4 4 2SiO Ca g 25 . 172Ca g 16 . 80 = 12 g Ca 12 g Ca3Al2O6 6 2 3O Al Ca g 20 . 270Ca g 24 . 120 = 5.3 g Ca 8.0 g Ca2AlFeO5 5 2AlFeO Ca g 99 . 242Ca g 16 . 80 = 2.6 g Ca 3.5 g CaSO4-2H2O O H 2 CaSO g 18 . 172Ca g 08 . 402 4 - = 0.81 g Ca Mass of Ca = 26 + 12 + 5.3 + 2.6 + 0.81 = 47 g Ca % Ca = cement g 0 . 100Ca g 47 100 = 47% Ca Mass percent Al: 12 g Ca3 Al2O6 6 2 3O Al Ca g 20 . 270Al g 96 . 53 = 2.4 g Al 8.0 g Ca2AlFeO5 5 2AlFeO Ca g 99 . 242Al g 98 . 26 = 0.89 g Al 24 CHAPTER 3 STOICHIOMETRY % Al = g 0 . 100g 89 . 0 g 4 . 2 + 100 = 3.3% Al Mass percent Fe: 8.0 g Ca2AlFeO5 5 2AlFeO Ca g 99 . 242Fe g 85 . 55= 1.8 g Fe; % Fe = g 0 . 100g 8 . 1 100 = 1.8% Fe Empirical and Molecular Formulas 43. a. Molar mass of CH2O = 1 mol C ||.|

\|C molg 011 . 12+ 2 mol H ||.|

\|H molg 0079 . 1 + 1 mol O ||.|

\|O molg 999 . 15= 30.026 g/mol % C = O CH g 026 . 30C g 011 . 122 100 = 40.002% C; % H = O CH g 026 . 30H g 0158 . 22 100 = 6.7135% H % O = O CH g 026 . 30O g 999 . 152 100 = 53.284% O or % O = 100.000 (40.002 + 6.7135) = 53.285% b. Molar mass of C6H12O6 = 6(12.011) + 12(1.0079) + 6(15.999) = 180.155 g/mol % C = 6 12 6O H C g 155 . 180C g 066 . 72 100 = 40.002%; % H = g 155 . 180g ) 0079 . 1 ( 12 100 = 6.7136% % O = 100.00 (40.002 + 6.7136) = 53.284% c. Molar Mass of HC2H3O2 = 2(12.011) + 4(1.0079) + 2(15.999) = 60.052 g/mol % C = g 052 . 60g 022 . 24 100 = 40.002%; % H = g 052 . 60g 0316 . 4 100 = 6.7135% % O = 100.000 (40.002 + 6.7135) = 53.285% All three compounds have the same empirical formula, CH2O, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula. 44. a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2. CHAPTER 3 STOICHIOMETRY 25 b. Molecular formula: C3H6; empirical formula: CH2 c. Molecular formula: P4O10; empirical formula: P2O5 d. Molecular formula: C6H12O6; empirical formula: CH2O 45. Compound I: mass O = 0.6498 g HgxOy 0.6018 g Hg = 0.0480 g O 0.6018 g Hg Hg g 6 . 200Hg mol 1 = 3.000 103 mol Hg 0.0480 g O O g 00 . 16O mol 1 = 3.00 103 mol O The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: mass Hg = 0.4172 g HgxOy 0.016 g O = 0.401 g Hg 0.401 g Hg Hg g 6 . 200Hg mol 1= 2.00 103 mol Hg; 0.016 g O O g 00 . 16O mol 1= 1.0 103 mol O The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 46. Out of 100.00 g of adrenaline, there are: 56.79 g C C g 011 . 12C mol 1 = 4.728 mol C; 6.56 g H H g 008 . 1H mol 1 = 6.51 mol H 28.37 g O O g 999 . 15O mol 1 = 1.773 mol O; 8.28 g N N g 01 . 14N mol 1 = 0.591 mol N Dividing each mole value by the smallest number: 591 . 0728 . 4 = 8.00; 591 . 051 . 6 = 11.0; 591 . 0773 . 1 = 3.00; 591 . 0591 . 0 = 1.00 This gives adrenaline an empirical formula of C8H11O3N. 47. First, we will determine composition in mass percent. We assume that all the carbon in the 0.213 g CO2 came from the 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O came from the 0.157 g of the compound. 0.213 g CO2 2CO g 01 . 44C g 01 . 12 = 0.0581 g C; % C = compound g 157 . 0C g 0581 . 0 100 = 37.0% C 26 CHAPTER 3 STOICHIOMETRY 0.0310 g H2O O H g 02 . 18H g 016 . 22 = 3.47 103 g H; % H = g 157 . 0g 10 47 . 33 100 = 2.21% H We get the mass percent of N from the second experiment: 0.0230 g NH3 3NH g 03 . 17N g 01 . 14 = 1.89 102 g N % N = g 103 . 0g 10 89 . 12 100 = 18.3% N The mass percent of oxygen is obtained by difference: % O = 100.00 (37.0 + 2.21 + 18.3) = 42.5% O So, out of 100.00 g of compound, there are: 37.0 g C C g 01 . 12C mol 1 = 3.08 mol C; 2.21 g H H g 008 . 1H mol 1 = 2.19 mol H 18.3 g N N g 01 . 14N mol 1 = 1.31 mol N; 42.5 g O O g 00 . 16O mol 1 = 2.66 mol O Lastly, and often the hardest part, we need to find simple whole number ratios. Divide all mole values by the smallest number: 31 . 108 . 3 = 2.35; 31 . 119 . 2 = 1.67; 31 . 131 . 1 = 1.00; 31 . 166 . 2 = 2.03 Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6. 48. Assuming 100.00 g of compound (mass oxygen = 100.00 g 41.39 g C 3.47 g H = 55.14 g O): 41.39 g C C g 011 . 12C mol 1 = 3.446 mol C; 3.47 g H H g 008 . 1H mol 1 = 3.44 mol H 55.14 g O O g 999 . 15O mol 1 = 3.446 mol O All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol. Molar mass = mol 129 . 0g 0 . 15 = 116 g/mol CHAPTER 3 STOICHIOMETRY 27 mass Empiricalmass Molar = 02 . 29116= 4.00; molecular formula = (CHO)4 = C4H4O4 49. Assuming 100.00 g of compound (mass hydrogen = 100.00 g 49.31 g C 43.79 g O = 6.90 g H): 49.31 g C C g 011 . 12C mol 1 = 4.105 mol C; 6.90 g H H g 008 . 1H mol 1 = 6.85 mol H 43.79 g O O g 999 . 15O mol 1 = 2.737 mol O Dividing all mole values by 2.737 gives: 737 . 2105 . 4 = 1.500; 737 . 285 . 6 = 2.50; 737 . 2737 . 2 = 1.000 Because a whole number ratio is required, the empirical formula is C3H5O2. Empirical formula mass - 3(12.0) + 5(1.0) +2(16.0) = 73.0 g/mol mass formula Empiricalmass Molar = 0 . 731 . 146 = 2.00; molecular formula = (C3H5O2)2 = C6H10O4 50. 41.98 mg CO2 2CO mg 009 . 44C mg 011 . 12 = 11.46 mg C; % C = mg 81 . 19mg 46 . 11 100 = 57.85% C 6.45 mg H2O O H mg 02 . 18H mg 016 . 22 = 0.722 mg H; % H = mg 81 . 19mg 722 . 0 100 = 3.64% H % O = 100.00 (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C C g 011 . 12C mol 1 = 4.816 mol C; 3.64 g H H g 008 . 1H mol 1 = 3.61 mol H 38.51 g O O g 999 . 15O mol 1 = 2.407 mol O 407 . 2816 . 4 = 2.001; 407 . 261 . 3 = 1.50; 407 . 2407 . 2 = 1.000 The C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2. 28 CHAPTER 3 STOICHIOMETRY Mass of C4H3O2 - 4(12) + 3(1) + 2(16) = 83 Molar mass = mol 250 . 0g 5 . 41 = 166 g/mol; 83166 = 2.0; the molecular formula is C8H6O4. 51. First, we will determine composition by mass percent: 16.01 mg CO2 gmg 1000CO g 009 . 44C g 011 . 12mg 1000g 12 = 4.369 mg C % C = compound mg 68 . 10C mg 369 . 4 100 = 40.91% C 4.37 mg H2O gmg 1000O H g 02 . 18H g 016 . 2mg 1000g 12 = 0.489 mg H % H = mg 68 . 10mg 489 . 0 100 = 4.58% H; % O = 100.00 - (40.91 + 4.58) = 54.51% O So, in 100.00 g of the compound, we have: 40.91 g C C g 011 . 12C mol 1 = 3.406 mol C; 4.58 g H H g 008 . 1H mol 1 = 4.54 mol H 54.51 g O O g 999 . 15O mol 1 = 3.407 mol O Dividing by the smallest number: 406 . 354 . 4 = 1.33 34; the empirical formula is C3H4O3. The empirical formula mass of C3H4O3 is - 3(12) + 4(1) + 3(16) = 88 g. Because 881 . 176 = 2.0, the molecular formula is C6H8O6. 52. a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N N H C mol 1N H C g 06 . 53N g 01 . 14N H C mol 13 33 3 3 3 = 33.3 g C3H3N % C3H3N = polymer g 00 . 100N H C g 3 . 333 3 = 33.3% C3H3N Only butadiene in the polymer reacts with Br2: 0.605 g Br2 6 46 426 422H C molH C g 09 . 54Br molH C mol 1Br g 8 . 159Br mol 1 = 0.205 g C4H6 CHAPTER 3 STOICHIOMETRY 29 % C4H6 = g 20 . 1g 205 . 0 100 = 17.1% C4H6 b. If we have 100.0 g of polymer: 33.3 g C3H3N g 06 . 53N H C mol 13 3 = 0.628 mol C3H3N 17.1 g C4H6 6 46 4H C g 09 . 54H C mol 1 = 0.316 mol C4H6 49.6 g C8H8 8 88 8H C g 14 . 104H C mol 1 = 0.476 mol C8H8 Dividing by 0.316: 316 . 0628 . 0 = 1.99; 316 . 0316 . 0 = 1.00; 316 . 0476 . 0 = 1.51 This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer, or (A4B2S3)n. Balancing Chemical Equations 53. When balancing reactions, start with elements that appear in only one of the reactants and one of the products, and then go on to balance the remaining elements. a. C6H12O6(s) + O2(g) CO2(g) + H2O(g) Balance C atoms: C6H12O6 + O2 6 CO2 + H2O Balance H atoms: C6H12O6 + O2 6 CO2 + 6 H2O Lastly, balance O atoms: C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) b. Fe2S3(s) + HCl(g) FeCl3(s) + H2S(g) Balance Fe atoms: Fe2S3 + HCl 2 FeCl3 + H2S Balance S atoms: Fe2S3 + HCl 2 FeCl3 + 3 H2S There are 6 H and 6 Cl on right, so balance with 6 HCl on left: Fe2S3(s) + 6 HCl(g) 2 FeCl3(s) + 3 H2S(g) c. CS2(l) + NH3(g) H2S(g) + NH4SCN(s) C and S are balanced; balance N: 30 CHAPTER 3 STOICHIOMETRY CS2 + 2 NH3 H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g) H2S(g) + NH4SCN(s). 54. An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. a. C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(g) b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq) Pb3(PO4)2(s) + 6 NaNO3(aq) 55. a. 16 Cr(s) + 3 S8(s) 8 Cr2S3(s) b. 2 NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) c. 2 KClO3(s) 2 KCl(s) + 3 O2(g) d. 2 Eu(s) + 6 HF(g) 2 EuF3(s) + 3 H2(g) e. 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g) 56. a. 2 KO2(s) + 2 H2O(l) 2 KOH(aq) + O2(g) + H2O2(aq) or 4 KO2(s) + 6 H2O(l) 4 KOH(aq) + O2(g) + 4 H2O2(aq) b. Fe2O3(s) + 6 HNO3(aq) 2 Fe(NO3)3(aq) + 3 H2O(l) c. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) d. PCl5(l) + 4 H2O(l) H3PO4(aq) + 5 HCl(g) e. 2 CaO(s) + 5 C(s) 2 CaC2(s) + CO2(g) f. 2 MoS2(s) + 7 O2(g) 2 MoO3(s) + 4 SO2(g) g. FeCO3(s) + H2CO3(aq) Fe(HCO3)2(aq) Reaction Stoichiometry 57. 1.000 kg Al 4 44 4 4 4ClO NH molClO NH g 49 . 117Al mol 3ClO NH mol 3Al g 98 . 26Al mol 1Al kgAl g 1000 = 4355 g NH4ClO4 58. 1.0 106 kg HNO3 3333HNO g 0 . 63HNO mol 1HNO kgHNO g 1000 = 1.6 107 mol HNO3 We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations: CHAPTER 3 STOICHIOMETRY 31 3223NH mol 4NO mol 4NO mol 2NO mol 2NO mol 3HNO mol 2 = 33NH mol 24HNO mol 16 Thus we can produce 16 mol HNO3 for every 24 mol NH3 that we begin with: 1.6 107 mol HNO3 3333NH molNH g 0 . 17HNO mol 16NH mol 24 = 4.1 108 g or 4.1 105 kg NH3 This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1 : 1; that is, 1 mol of NH3 produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 105 kg NH3 reacted. 59. Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s) 15.0 g Fe Fe g 85 . 55Fe mol 1= 0.269 mol Fe; 0.269 mol Fe Al molAl g 98 . 26Fe mol 2Al mol 2 = 7.26 g Al 0.269 mol Fe 3 23 2 3 2O Fe molO Fe g 70 . 159Fe mol 2O Fe mol 1 = 21.5 g Fe2O3 0.269 mol Fe 3 23 2 3 2O Al molO Al g 96 . 101Fe mol 2O Al mol 1 = 13.7 g Al2O3 60. 10 KClO3(s) + 3 P4(s) 3 P4O10(s) + 10 KCl(s) 52.9 g KClO3 10 410 4310 433O P molO P g 88 . 283KClO mol 10O P mol 3KClO g 55 . 122KClO mol 1 = 36.8 g P4O10 61. 2 LiOH(s) + CO2(g) Li2CO3(aq) + H2O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 20. = 140 L/min. 25,000 g LiOH 2 22 2CO g 0 . 4air g 100CO molCO g 01 . 44LiOH mol 2CO mol 1LiOH g 95 . 23LiOH mol 1 min 60h 1air L 140min 1mL 1000L 1air g 0010 . 0air mL 1 = 68 h = 2.8 days 62. 1.0 104 kg waste + ++ + 42 7 544 4NH mol 55N O H C mol 1NH g 04 . 18NH mol 1kgg 1000waste kg 100NH kg 0 . 3 N O H C molN O H C g 1 . 1132 7 52 7 5 = 3.4 104 g tissue if all NH4+ converted 32 CHAPTER 3 STOICHIOMETRY Because only 95% of the NH4+ ions react: mass of tissue = (0.95)(3.4 104 g) = 3.2 104 g or 32 kg bacterial tissue 63. 1.0 103 g phosphorite 2 4 32 4 3 2 4 3) PO ( Ca g 18 . 310) PO ( Ca mol 1e phosphorit g 100) PO ( Ca g 75 2 4 34) PO ( Ca mol 2P mol 1 44P molP g 88 . 123= 150 g P4 64. Total mass of copper used: 10,000 boards 3cmg 96 . 8board) cm 060 . 0 cm 0 . 16 cm 0 . 8 ( = 6.9 105 g Cu Amount of Cu to be recovered = 0.80 (6.9 105 g) = 5.5 105 g Cu 5.5 105 g Cu 2 4 32 4 3 2 4 3Cl ) NH ( Cu molCl ) NH ( Cu g 6 . 202Cu molCl ) NH ( Cu mol 1Cu g 55 . 63Cu mol 1 = 1.8 106 g Cu(NH3)4Cl2 5.5 105 g Cu 33 3NH molNH g 03 . 17Cu molNH mol 4Cu g 55 . 63Cu mol 1 = 5.9 105 g NH3 Limiting Reactants and Percent Yield 65. The product formed in the reaction is NO2; the other species present in the product represent-tation is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react with 3 O2 molecules to form 6 NO2 molecules. 6 NO(g) + 3 O2(g) 6 NO2(g) For smallest whole numbers, the balanced reaction is: 2 NO(g) + O2(g) 2 NO2(g) 66. In the following table we have listed three rows of information. The Initial row is the number of molecules present initially, the Change row is the number of molecules that react to reach completion, and the Final row is the number of molecules present at completion. To determine the limiting reactant, lets calculate how much of one reactant is necessary to react with the other. 10 molecules O2 23O molecules 5NH molecules 4 = 8 molecules NH3 to react with all the O2 CHAPTER 3 STOICHIOMETRY 33 Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Initial 10 molecules 10 molecules 0 0 Change 8 molecules 10 molecules +8 molecules +12 molecules Final 2 molecules 0 8 molecules 12 molecules The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules. 67. 1.50 g BaO2 22BaO g 3 . 169BaO mol 1 = 8.86 103 mol BaO2 25.0 mLHCl g 46 . 36HCl mol 1mLHCl g 0272 . 0 = 1.87 102 mol HCl The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual mole ratio is: 232BaO mol 10 86 . 8HCl mol 10 87 . 1 = 2.11 Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent. 8.86 103 mol BaO2 2 22 222 2O H molO H g 02 . 34BaO molO H mol 1 = 0.301 g H2O2 The amount of HCl reacted is: 8.86 103 mol BaO2 2BaO molHCl mol 2 = 1.77 102 mol HCl Excess mol HCl = 1.87 102 mol 1.77 102 mol = 1.0 103 mol HCl Mass of excess HCl = 1.0 103 mol HCl HCl molHCl g 46 . 36 = 3.6 102 g HCl 68. 25.0 g Ag2O g 8 . 231mol 1 = 0.108 mol Ag2O 50.0 g C10H10N4SO2 g 29 . 250mol 1 = 0.200 mol C10H10N4SO2 34 CHAPTER 3 STOICHIOMETRY O Ag MolSO N H C Mol22 4 10 10(actual) = 108 . 0200 . 0 = 1.85 The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting. 0.200 mol C10H10N4SO2 2 4 9 10 2 4 10 102 4 9 10SO N H AgC molg 357.18SO N H C mol 2SO N H AgC mol 2 = 71.4 g AgC10H9N4SO2 produced 69. 2.50 metric tons Cu3FeS3 3 33 3FeS Cu mol 1Cu mol 3g 71 . 342FeS Cu mol 1kgg 1000ton metrickg 1000 Cu molg 55 . 63 = 1.39 106 g Cu (theoretical) 1.39 106 g Cu (theoretical) ) l theoretica ( Cu g . 100) actual ( Cu g 3 . 86 = 1.20 106 g Cu = 1.20 103 kg Cu = 1.20 metric tons Cu (actual) 70. a. 1142 g C6H5Cl Cl H C g 55 . 112Cl H C mol 15 65 6= 10.15 mol C6H5Cl 485 g C2HOCl3 3 23 2HOCl C g 38 . 147HOCl C mol 1 = 3.29 mol C2HOCl3 From the balanced equation, the required mole ratio is 3 25 6HOCl C mol 1Cl H C mol 2= 2. The actual mole ratio present is 3 25 6HOCl C mol 29 . 3Cl H C mol 15 . 10= 3.09. The actual mole ratio is greater than the required mole ratio, so the denominator of the actual mole ratio (C2HOCl3) is limiting. 3.29 mol C2HOCl3 5 9 145 9 143 25 9 14Cl H C molCl H C g 46 . 354HOCl C molCl H C mol 1 = 1170 g C14H9Cl5 (DDT) b. C2HOCl3 is limiting, and C6H5Cl is in excess. c. 3.29 mol C2HOCl3 Cl H C molCl H C g 55 . 112HOCl C molCl H C mol 25 65 63 25 6 = 741 g C6H5Cl reacted 1142 g 741 g = 401 g C6H5Cl in excess d. Percent yield = DDT g 1170DDT g 0 . 200 100 = 17.1% CHAPTER 3 STOICHIOMETRY 35 71. An alternative method to solve limiting-reagent problems is to assume that each reactant is limiting and then calculate how much product could be produced from each reactant. The reactant that produces the smallest amount of product will run out first and is the limiting reagent. 5.00 106 g NH3 3 33NH mol 2HCN mol 2NH g 03 . 17NH mol 1 = 2.94 105 mol HCN 5.00 106 g O2 2 22O mol 3HCN mol 2O g 00 . 32O mol 1 = 1.04 105 mol HCN 5.00 106 g CH4 4 44CH mol 2HCN mol 2CH g 04 . 16CH mol 1 = 3.12 105 mol HCN O2 is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH3 and CH4, only enough O2 is present to produce 1.04 105 mol HCN. The mass of HCN that can be produced is: 1.04 105 mol HCN HCN molHCN g 03 . 27 = 2.81 106 g HCN 5.00 106 g O2 O H mol 1O H g 02 . 18O mol 3O H mol 6O g 00 . 32O mol 1222222 = 5.63 106 g H2O 72. 2 C3H6(g) + 2 NH3(g) + 3 O2(g) 2 C3H3N(g) + 6 H2O(g) a. We will solve this limiting reagent problem using the same method as described in Exercise 3.71. 1.00 103 g C3H6 6 33 36 36 3H C mol 2N H C mol 2H C g 08 . 42H C mol 1 = 23.8 mol C3H3N 1.50 103 g NH3 33 333NH mol 2N H C mol 2NH g 03 . 17NH mol 1 = 88.1 mol C3H3N 2.00 103 g O2 23 322O mol 3N H C mol 2O g 00 . 32O mol 1 = 41.7 mol C3H3N C3H6 is limiting because it produces the smallest amount of product, and the mass of acrylonitrile that can be produced is: 23.8 mol molN H C g 06 . 533 3 = 1.26 103 g C3H3N 36 CHAPTER 3 STOICHIOMETRY b. 23.8 mol C3H3N O H molO H g 02 . 18N H C mol 2O H mol 6223 32 = 1.29 103 g H2O Amount NH3 needed: 23.8 mol C3H3N 333 33NH molNH g 03 . 17N H C mol 2NH mol 2 = 405 g NH3 Amount NH3 in excess = 1.50 103 g 405 g = 1.10 103 g NH3 Amount O2 needed: 23.8 mol C3H3N 223 32O molO g 00 . 32N H C mol 2O mol 3 = 1.14 103 g O2 Amount O2 in excess = 2.00 103 g 1.14 103 g = 860 g O2 1.10 103 g NH3 and 860 g O2 are in excess. 73. P4(s) + 6 F2(g) 4 PF3(g); the theoretical yield of PF3 is: 120. g PF3 (actual) ) actual ( PF g 1 . 78) l theoretica ( PF g 0 . 10033 = 154 g PF3 (theoretical) 154 g PF3 223233F molF g 00 . 38PF mol 4F mol 6PF g 97 . 87PF mol 1 = 99.8 g F2 99.8 g F2 are needed to actually produce 120. g of PF3 if the percent yield is 78.1%. 74. a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every 4 mol of C6H5O3N reacted. The actual yield is 3 mol of acetaminophen compared with a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen): percent yield = M mol 4M mol 3 100 = 75% b. The product of the percent yields of the individual steps must equal the overall yield, 75%. (0.87)(0.98)(x) = 0.75, x = 0.88; step III has a percent yield of 88%. CHAPTER 3 STOICHIOMETRY 37 Additional Exercises 75. 17.3 g H H g 008 . 1H mol 1 = 17.2 mol H; 82.7 g C C g 01 . 12C mol 1 = 6.89 mol C 89 . 62 . 17 = 2.50; the empirical formula is C2H5. The empirical formula mass is ~29 g, so two times the empirical formula would put the compound in the correct range of the molar mass. Molecular formula = (C2H5)2 = C4H10 2.59 1023 atoms H molecules 10 022 . 6H C mol 1H atoms 10H C molecule 12310 4 10 4 = 4.30 102 mol C4H10 4.30 102 mol C4H10 10 4H C molg 12 . 58 = 2.50 g C4H10 76. Assuming 100.00 g E3H8: Mol E = 8.73 g H H mol 8E mol 3H g 008 . 1H mol 1 = 3.25 mol E E mol 1E g x = E mol 25 . 3E g 27 . 91, x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 amu 77. Molar mass X2 = molecules 10 022 . 6mol 1molecules 10 92 . 8g 105 . 02320 = 70.9 g/mol The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine. Assuming 100.00 g of MX3 compound: 54.47 g Cl g 45 . 35mol 1 = 1.537 mol Cl 1.537 mol Cl Cl mol 3M mol 1 = 0.5123 mol M Molar mass of M = M mol 5123 . 0M g 53 . 45 = 88.87 g/mol M M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride. The balanced equation is 2 Y + 3 Cl2 2 YCl3. Assuming Cl2 is limiting: 38 CHAPTER 3 STOICHIOMETRY 1.00 g Cl2 332322YCl mol 1YCl g 26 . 195Cl mol 3YCl mol 2Cl g 90 . 70Cl mol 1 = 1.84 g YCl3 Assuming Y is limiting: 1.00 g Y 33 3YCl mol 1YCl g 26 . 195Y mol 2YCl mol 2Y g 91 . 88Y mol 1 = 2.20 g YCl3 Because Cl2, when it all reacts, produces the smaller amount of product, Cl2 is the limiting reagent, and the theoretical yield is 1.84 g YCl3. 78. Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; because 104.14/13.02 = 7.998 - 8, the molecular formula for styrene is (CH)8 = C8H8. 2.00 g C8H8 H molatoms 10 022 . 6H C molH mol 8H C g 14 . 104H C mol 1238 8 8 88 8 = 9.25 1022 atoms H 79. Mass of H2O = 0.755 g CuSO4-xH2O 0.483 g CuSO4 = 0.272 g H2O 0.483 g CuSO4 44CuSO g 62 . 159CuSO mol 1 = 0.00303 mol CuSO4 0.272 g H2O O H g 02 . 18O H mol 122 = 0.0151 mol H2O 4242CuSO mol 1O H mol 98 . 4CuSO mol 00303 . 0O H mol 0151 . 0= ; compound formula = CuSO4-5H2O, x = 5 80. In 1 hour, the 1000. kg of wet cereal contains 580 kg H2O and 420 kg of cereal. We want the final product to contain 20.% H2O. Let x = mass of H2O in final product. xx+ 420= 0.20, x = 84 + (0.20)x, x = 105 - 110 kg H2O The amount of water to be removed is 580 110 = 470 kg/h. 81. Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2 anions. The simplest compound of the two elements is Al2O3. Similarly, we would expect the formula of a Group 6A element with Al to be Al2X3. Assuming this, out of 100.00 g of compound, there are 18.56 g Al and 81.44 g of the unknown element, X. Lets use this information to determine the molar mass of X, which will allow us to identify X from the periodic table. 18.56 g Al Al mol 2X mol 3Al g 98 . 26Al mol 1 = 1.032 mol X 81.44 g of X must contain 1.032 mol of X. CHAPTER 3 STOICHIOMETRY 39 Molar mass of X = X mol 032 . 1X g 44 . 81 = 78.91 g/mol From the periodic table, the unknown element is selenium, and the formula is Al2Se3. 82. The reaction is BaX2(aq) + H2SO4(aq) BaSO4(s) + 2 HX(aq). 0.124 g BaSO4 4BaSO g 4 . 233Ba g 3 . 137 = 0.0729 g Ba; % Ba = 2BaX g 158 . 0Ba g 0729 . 0 100 = 46.1% The formula is BaX2 (from positions of the elements in the periodic table), and 100.0 g of compound contains 46.1 g Ba and 53.9 g of the unknown halogen. There must also be: 46.1 g Ba Ba molX mol 2Ba g 3 . 137Ba mol 1 = 0.672 mol of the halogen in 53.9 g of halogen Therefore, the molar mass of the halogen is mol 672 . 0g 9 . 53 = 80.2 g/mol. This molar mass is close to that of bromine. Thus the formula of the compound is BaBr2. 83. 1.20 g CO2 O N H C molg 51 . 376C mol 24O N H C mol 1CO molC mol 1g 01 . 44CO mol 13 30 243 30 2422 = 0.428 g C24H30N3O sample g 00 . 1O N H C g 428 . 03 30 24 100 = 42.8% C24H30N3O (LSD) 84. Ca3(PO4)2(s) + 3 H2SO4(aq) 3 CaSO4(s) + 2 H3PO4(aq) 1.0 103 g Ca3(PO4)2 2 4 32 4 3) PO ( Ca g 2 . 310) PO ( Ca mol 1 = 3.2 mol Ca3(PO4)2 1.0 103 g conc. H2SO4 4 24 24 24 2SO H g 1 . 98SO H mol 1SO H . conc g 100SO H g 98 = 10. mol H2SO4 The required mole ratio from the balanced equation is 3 mol H2SO4 to 1 mol Ca3(PO4)2. The actual ratio is 2 4 34 2) PO ( Ca mol 2 . 3SO H mol . 10 = 3.1. This is larger than the required mole ratio, so Ca3(PO4)2 is the limiting reagent. 3.2 mol Ca3(PO4)2 442 4 34CaSO molCaSO g 2 . 136) PO ( Ca molCaSO mol 3 = 1300 g CaSO4 produced 40 CHAPTER 3 STOICHIOMETRY 3.2 mol Ca3(PO4)2 4 34 32 4 34 3PO H molPO H g 0 . 98) PO ( Ca molPO H mol 2 = 630 g H3PO4 produced 85. 453 g Fe 3 23 2 3 2O Fe molO Fe g 70 . 159Fe mol 2O Fe mol 1Fe g 85 . 55Fe mol 1 = 648 g Fe2O3 Mass % Fe2O3 = ore g 752O Fe g 6483 2 100 = 86.2% 86. 12C21H6: 2(12.000000) + 6(1.007825) = 30.046950 amu 12C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 amu 14N16O: 1(14.003074) + 1(15.994915) = 29.997989 amu The peak results from 12C1H216O. 87. atoms Rbatoms Rb8785 = 2.591; If we had exactly 100 atoms, x = number of 85Rb atoms and 100 x = number of 87Rb atoms. xx 100 = 2.591, x = 259.1 (2.591)x, x = 591 . 31 . 259 = 72.15; 72.15% 85Rb 0.7215(84.9117) + 0.2785(A) = 85.4678, A = 2785 . 026 . 61 4678 . 85 = 86.92 amu 88. Assuming 100.00 g of tetrodotoxin: 41.38 g C C g 011 . 12C mol 1 = 3.445 mol C; 13.16 g N N g 007 . 14N mol 1 = 0.9395 mol N 5.37 g H H g 008 . 1H mol 1 = 5.33 mol H; 40.09 g O O g 999 . 15O mol 1 = 2.506 mol O Divide by the smallest number: 9395 . 0445 . 3 = 3.667; 9395 . 033 . 5 = 5.67; 9395 . 0506 . 2 = 2.667 To get whole numbers for each element, multiply through by 3. Empirical formula: (C3.667H5.67NO2.667)3 = C11H17N3O8; the mass of the empirical formula is 319.3 g/mol. CHAPTER 3 STOICHIOMETRY 41 Molar mass tetrodotoxin = molecules 10 022 . 6mol 1molecules 3g 10 59 . 12321 = 319 g/mol Because the empirical mass and molar mass are the same, the molecular formula is the same as the empirical formula, C11H17N3O8. 165 lb mol 1molecules 10 022 . 6g 3 . 319mol 1g g 10 1kgg . 10lb 2046 . 2kg 123 6 = 1.4 1018 molecules tetrodotoxin is the LD50 dosage 89. The volume of a gas is proportional to the number of molecules of gas. Thus the formulas are: I: NH3 II: N2H4 III: HN3 The mass ratios are: I: H gN g 634 . 4 II: H gN g 949 . 6 III: H gN g 7 . 41 If we set the atomic mass of H equal to 1.008, then the atomic mass for nitrogen is: I: 14.01 II: 14.01 III. 14.0 For example, for compound I: ) 008 . 1 ( 3A = 1634 . 4, A = 14.01 90. PaO2 + O2 PaxOy (unbalanced) 0.200 g PaO2 2PaO g 263Pa g 231 = 0.1757 g Pa (We will carry an extra significant figure.) 0.2081 g PaxOy 0.1757 g Pa = 0.0324 g O; 0.0324 g O O g 00 . 16O mol 1 = 2.025 103 mol O 0.1757 g Pa Pa g 231Pa mol 1 = 7.61 104 mol Pa Pa MolO Mol = Pa mol 10 61 . 7O mol 10 025 . 243 = 2.66 Pa mol 3O mol 8322 = ; empirical formula: Pa3O8 91. 1.375 g AgI AgI g 8 . 234AgI mol 1 = 5.856 103 mol AgI = 5.856 103 mol I 1.375 g AgI AgI g 8 . 234I g 9 . 126 = 0.7431 g I; XI2 contains 0.7431 g I and 0.257 g X. 42 CHAPTER 3 STOICHIOMETRY 5.856 103 mol I I mol 2X mol 1 = 2.928 103 mol X Molar mass = molg 8 . 87X mol 10 928 . 2X g 257 . 03 =; atomic mass = 87.8 amu (X is Sr.) 92. X2Z: 40.0% X and 60.0% Z by mass; xzzxA ) 0 . 60 (A ) 0 . 40 (60.0/A0.0/A 42Z molX mol= = = or Az = 3Ax where A = molar mass For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax. Mass % X = xxA 7A 100 = 14.3% X; % Z = 100.0 14.3 = 85.7% Z 93. Assuming 1 mole of vitamin A (286.4 g vitamin A): mol C = 286.4 g vitamin A C g 12.011C mol 1A vitamin gC g 0.8396 = 20.00 mol C mol H = 286.4 g vitamin A H g 1.0079H mol 1A vitamin gH g 0.1056 = 30.01 mol H Because 1 mole of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, lets calculate the molar mass of E: 286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol From the periodic table, E = oxygen, and the molecular formula of vitamin A is C20H30O. 94. We would see the peaks corresponding to: 10B35Cl3 [mass - 10 + 3(35) = 115 amu], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119), 10B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122) We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122. Challenge Problems 95. When the discharge voltage is low, the ions present are in the form of molecules. When the discharge voltage is increased, the bonds in the molecules are broken, and the ions present are in the form of individual atoms. Therefore, the high discharge data indicate that the ions 16O+, 18O+, and 40Ar+ are present. The only combination of these individual ions that can explain the mass data at low discharge is 16O16O+ (mass = 32), 16O18O+ (mass = 34), and 40Ar+ (mass = 40). Therefore, the gas mixture contains 16O16O, 16O18O, and 40Ar. To determine the CHAPTER 3 STOICHIOMETRY 43 percent composition of each isotope, we use the relative intensity data from the high discharge data to determine the percentage that each isotope contributes to the total relative intensity. For 40Ar: = = + +1007515 . 10000 . 11000000 . 1 0015 . 0 7500 . 00000 . 157.094% 40Ar For 16O: 7515 . 17500 . 0 100 = 42.82% 16O; for 18O: 7515 . 10015 . 0 100 = 8.6 102% 18O Note: 18F instead of 18O could also explain the data. However, OF(g) is not a stable compound. This is why 18O is the best choice because O2(g) does form. 96. Fe(s) + ) s ( FeO ) g ( O221 ; 2 Fe(s) + ) s ( O Fe ) g ( O3 2 223 20.00 g Fe g 85 . 55Fe mol 1 = 0.3581 mol g 00 . 32O mol 1O g ) 24 . 3 20 . 11 (22 = 0.2488 mol O2 consumed (1 extra sig. fig.) Assuming x mol of FeO is produced from x mol of Fe, so that 0.3581 x mol of Fe reacts to form Fe2O3: x Fe + FeO O212x x 3 2 2O Fe mol23581 . 0O mol23581 . 023Fe mol ) 3581 . 0 ( |.|

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\| + x xx Setting up an equation for total moles of O2 consumed: FeO mol 079 . 0 0791 . 0 , O mol 2488 . 0 ) 3581 . 0 (24321= = = + x x x 0.079 mol FeO molFeO g 85 . 71 = 5.7 g FeO produced Mol Fe2O3 produced = 2079 . 0 3581 . 0 = 0.140 mol Fe2O3 0.140 mol Fe2O3 molO Fe g 70 . 1593 2 = 22.4 g Fe2O3 produced 97. 10.00 g XCl2 + excess Cl2 12.55 g XCl4; 2.55 g Cl reacted with XCl2 to form XCl4. XCl4 contains 2.55 g Cl and 10.00 g XCl2. From mole ratios, 10.00 g XCl2 must also contain 2.55 g Cl; mass X in XCl2 = 10.00 2.55 = 7.45 g X. CHAPTER 3 STOICHIOMETRY 44 2.55 g Cl 22XCl molX mol 1Cl mol 2XCl mol 1Cl g 45 . 35Cl mol 1 = 3.60 210mol X So, 3.60 210mol X has a mass equal to 7.45 g X. The molar mass of X is: X mol 10 60 . 3X g 45 . 72 = 207 g/mol X; atomic mass = 207 amu, so X is Pb. 98. The two relevant equations are: Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) Let x = mass Mg, so 10.00 x = mass Zn. From the balanced equations, moles H2 = moles Zn + moles Mg. Mol H2 = 0.5171 g H2 22H g 0158 . 2H mol 1 = 0.2565 mol H2 0.2565 = 31 . 24x + 38 . 6500 . 10 x ; solving, x = 4.008 g Mg. g 00 . 10g 008 . 4 100 = 40.08% Mg 99. For a gas, density and molar mass are directly proportional to each other. Molar mass XHn = 2.393(32.00) = molg 58 . 76; 0.803 g H2O O H g 02 . 18H mol 22 = 8.91 102 mol H n nXH molH mol 4XH mol 10 2.23H mol 10 8.9122= Molar mass X = 76.58 4(1.008 g) = 72.55 g/mol; the element is Ge. 100. The balanced equation is 2 Sc(s) + 2x HCl(aq) 2 ScClx(aq)+ x H2(g). The mol ratio of Sc to H2 = x2. Mol Sc = 2.25 g Sc Sc g 96 . 44Sc mol 1 = 0.0500 mol Sc Mol H2 = 0.1502 g H2 22H g 0158 . 2H mol 1 = 0.07451 mol H2 x2 = 07451 . 00500 . 0, x = 3; the formula is ScCl3. CHAPTER 3 STOICHIOMETRY 45 101. 4.000 g M2S3 3.723 g MO2 There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M: 00 . 32 A723 . 321 . 96 A 2000 . 8,) 00 . 16 ( 2 Ag 723 . 3) 07 . 32 ( 3 A 2g 000 . 42+ + +||.|

\|+ = = (8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A = 102.2, A = 184 g/mol; atomic mass = 184 amu 102. We know that water is a product, so one of the elements in the compound is hydrogen. XaHb + O2 H2O + ? To balance the H atoms, the mole ratio between XaHb and H2O = b2. Mol compound = mol / g 09 . 62g 39 . 1 = 0.0224 mol; mol H2O = mol / g 02 . 18g 21 . 1 = 0.0671 mol 0671 . 00224 . 0 2=b, b = 6; XaH6 has a molar mass of 62.09 g/mol. 62.09 = a(molar mass of X) + 6(1.008), a(molar mass of X ) = 56.04 Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), and Li (a = 8). N fits the data best so N4H6 is the most likely formula. 103. The balanced equations are: C(s) + 1/2 O2(g) CO(g) and C(s) + O2(g) CO2(g) If we have 100.0 mol of products, then we have 72.0 mol CO2, 16.0 mol CO, and 12.0 mol O2. The initial moles of C equals 72.0 (from CO2) + 16.00 (from CO) = 88.0 mol C and the initial moles of O2 equals 72.0 (from CO2) + 16.0/2 (from CO) + 12.0 (unreacted O2) = 92.0 mol O2. The initial reaction mixture contained: C mol 0 . 88O mol 0 . 922 = 1.05 mol O2/mol C 104. CxHyOz + oxygen x CO2 + y/2 H2O Mass % C in aspirin = aspirin g 00 . 1C molC g 01 . 12CO molC mol 1CO g 01 . 44CO mol 1CO g 20 . 22 222 = 60.0% C 46 CHAPTER 3 STOICHIOMETRY Mass % H in aspirin = aspirin g 00 . 1H molH g 008 . 1O H molH mol 2O H g 02 . 18O H mol 1O H g 400 . 02 222 = 4.48% H Mass % O = 100.00 (60.0 + 4.48) = 35.5% O Assuming 100.00 g aspirin: 60.0 g C C g 01 . 12C mol 1 = 5.00 mol C; 4.48 g H H g 008 . 1H mol 1 = 4.44 mol H 35.5 g O O g 00 . 16O mol 1 = 2.22 mol O Dividing by the smallest number: 22 . 200 . 5 = 2.25; 22 . 244 . 4 = 2.00 Empirical formula: (C2.25 H2.00O)4 = C9H8O4. Empirical mass ~ 9(12) + 8(1) + 4(16) = 180 g/mol; this is in the 170190 g/mol range, so the molecular formula is also C9H8O4. Balance the aspirin synthesis reaction to determine the formula for salicylic acid. CaHbOc + C4H6O3 C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3 105. LaH2.90 is the formula. If only La3+ is present, LaH3 would be the formula. If only La2+ is present, LaH2 would be the formula. Let x = mol La2+ and y = mol La3+: (La2+)x(La3+)yH(2x + 3y) where x + y = 1.00 and 2x + 3y = 2.90 Solving by simultaneous equations: 2x + 3y = 2.90 2x 2y = 2.00 y = 0.90 and x = 0.10 LaH2.90 contains 101La2+, or 10.% La2+, and 109La3+, or 90.% La3+. 106. 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) 30.07 g/mol 44.09 g/mol Let x = mass C2H6, so 9.780 x = mass C3H8. Use the balanced reaction to set up an equation for the moles of O2 required. 1509 . 44780 . 92707 . 30 + x x = 1.120 mol O2 CHAPTER 3 STOICHIOMETRY 47 Solving: x = 3.7 g C2H6; g 780 . 9g 7 . 3 100 = 38% C2H6 by mass 107 . Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g. Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol Mol KCl = 55 . 74x; mol KNO3 = 11 . 101y Knowing that the mixture is 43.2% K, then in the 100.0 g mixture: 39.10 |.|

\| +11 . 101 55 . 74y x = 43.2 We have two equations and two unknowns: (0.5245)x + (0.3867)y = 43.2 x + y = 100.0 Solving, x = 32.9 g KCl; g 0 . 100g 9 . 32 100 = 32.9% KCl 108. Let M = unknown element Mass % M = 708 . 3077 . 2100compound mass totalM mass= 100 = 56.01% M 100.00 56.01 = 43.99% O Assuming 100.00 g compound: 43.99 g O O g 999 . 15O mol 1 = 2.750 mol O If MO is the formula of the oxide, then M has a molar mass of M mol 750 . 2M g 01 . 56 = 20.37 g/mol. This is too low for the molar mass. We must have fewer moles of M than moles O present in the formula. Some possibilities are MO2, M2O3, MO3, etc. It is a guessing game as to which to try. Lets assume an MO2 formula. Then the molar mass of M is: O mol 2M mol 1O mol 750 . 2M g 01 . 56 = 40.73 g/mol This is close to calcium, but calcium forms an oxide having the CaO formula, not CaO2. 48 CHAPTER 3 STOICHIOMETRY If MO3 is assumed to be the formula, then the molar mass of M calculates to be 61.10 g/mol which is too large. Therefore, the mol O to mol M ratio must be between 2 and 3. Some reasonable possibilities are 2.25, 2.33, 2.5, 2.67, and 2.75 (these are reasonable because they will lead to whole number formulas). Trying a mol O to mol M ratio of 2.5 to 1 gives a molar mass of: O mol 5 . 2M mol 1O mol 750 . 2M g 01 . 56 = 50.92 g/mol This is the molar mass of vanadium and V2O5 is a reasonable formula for an oxide of vanadium. The other choices for the O : M mole ratios between 2 and 3 do not give as reasonable results. Therefore, M is vanadium, and the formula is V2O5. 109. The balanced equations are: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(g) Let 4x = number of moles of NO formed, and let 4y = number of moles of NO2 formed. Then: 4x NH3 + 5x O2 4x NO + 6x H2O and 4y NH3 + 7y O2 4y NO2 + 6y H2O All the NH3 reacted, so 4x + 4y = 2.00. 10.00 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25. Solving by the method of simultaneous equations: 20x + 28y = 13.0 20x 20y = 10.0 8y = 3.0, y = 0.38; 4x + 4 0.38 = 2.00, x = 0.12 Mol NO = 4x = 4 0.12 = 0.48 mol NO formed 110. a N2H4 + b NH3 + (10.00 4.062) O2 c NO2 + d H2O Setting up four equations to solve for the four unknowns: 2a + b = c (N mol balance) 2c + d = 2(10.00 4.062) (O mol balance) 4a + 3b = 2d (H mol balance) a(32.05) + b(17.03) = 61.00 (mass balance) Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4. g 00 . 61H N mol / g 05 . 32 H N mol 1 . 14 2 4 2 100 = 58% N2H4 CHAPTER 3 STOICHIOMETRY 49 Marathon Problems 111. To solve the limiting-reagent problem, we must determine the formulas of all the com